The Weierstrass Form

Using Bezout’s Theorem, it can be shown that every irreducible cubic
has a flex
(a point where the tangent intersects the curve with multiplicity three)
or a singular point (a point where there is no tangent because both partial
derivatives are zero). [Reducible cubics consist of a line and a conic, which
are easy to study.]

Irreducible cubics containing singular points can be affinely transformed
into one of the following forms:

$Y^2 = X^3$

$Y^2 = X^2(X-1)$

$Y^2 = X^2(X+1)$

An irreducibe cubic with a flex can be affinely
transformed into a Weierstrass equation:

Fact: Isomorphic curves over some field $K$ have the same $j$-invariant.
Two curves with the same $j$-invariant are isomorphic over $\bar {K}$.

This equation can be further simplified through another affine transformation.
Let $K$ denote the field we are working in.
If the $\mathrm{char} K \ne 2$, then completing the square
on the left hand side (and performing an appropriate variable substitution)
eliminates the $XY$ and $Y$ terms. In addition,
if $\mathrm{char} K \ne 3$, then a similar trick eliminates
the $X^2$ term (whereas if $\mathrm{char} K = 3$ we can eliminate either the $X^2$
or the $X$ term).

If $\mathrm{char} K = 2$ then one of the following two forms can be obtained:

$Y^2 + XY = X^3 + a_2 X^2 + a_6$ (the nonsupersingular case)

$Y^2 + a_3 Y = X^3 + a_4 X + a_6$ (the supersingular case)

In Weierstrass form, we see that for any given value of $X$, there are at most
two values that $Y$ may take. If $a_1 = a_3 = 0$ (which is always the case
for $\mathrm{char} K \ne 2$, we have that if $(x,y)$ is a point, then $(x, -y)$ is
the other point with the same $x$-coordinate.

Example

Consider the cubic Fermat curve

\[
X^3 + Y^3 = 1
\]

Assume $\mathrm{char} K \ne 3$ (otherwise the curve is the same as $(X + Y)^3 = 1$).
An affine transformation takes it to its Weierstrass form: